codeforces825 D. Suitable Replacement 二分答案

D. Suitable Replacement

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

You are given two strings s and
t consisting of small Latin letters, string
s can also contain '?' characters.

Suitability of string
s is calculated by following metric:

Any two letters can be swapped positions, these operations can be performed arbitrary number of times over any pair of positions. Among all resulting strings
s, you choose the one with the largest number of
non-intersecting occurrences of string
t. Suitability is this number of occurrences.

You should replace all '?' characters with small Latin letters in such a way that the
suitability of string
s is maximal.

Input
The first line contains string s (1 ≤ |s| ≤ 106).

The second line contains string t (1 ≤ |t| ≤ 106).

Output
Print string s with
'?' replaced with small Latin letters in such a way that
suitability of that string is maximal.

If there are multiple strings with maximal suitability then print any of them.

Examples

Input

?aa?
ab

Output

baab

Input

??b?
za

Output

azbz

Input

abcd
abacaba

Output

abcd

Note
In the first example string "baab" can be transformed to
"abab" with swaps, this one has
suitability of 2. That means that string
"baab" also has
suitability of 2.

In the second example maximal suitability you can achieve is
1 and there are several dozens of such strings,
"azbz" is just one of them.

In the third example there are no '?' characters and the
suitability of the string is
0.

题意：给S ,T两个字符串，让你输出一个和S相同长度的串，并将S中的？改为小写字母，满足输出串中T串数量最多

思路：二分答案，模拟太烦人，搞得我乱死了（主要是太弱），先二分出最多有几个T串，然后输出的时候判断S串中哪个字母不够就输出，最后全够了就随便输出一个字母。

还有注意long long，我的一下午贡献给这题也是没谁了

#include <bits/stdc++.h>
#define LL long long
using namespace std;
const int AX = 1e6+66;
char a[AX];
char b[AX];
int numa[27];
int numb[27];
int num;
int lena,lenb;
bool check(int x){ //不用long long wa到死
LL c=0;
for(int i = 0 ; i < 26 ; i ++ )
if( numb[i] ){
c += max(0LL,1LL*x*numb[i]-numa[i]);
}
return c <= num;
}
int main(){
num = 0;
gets(a);
gets(b);
lena = strlen(a);
lenb = strlen(b);
for( int i = 0 ; i < lena ; i++ ){
if( a[i] >= 'a' && a[i] <= 'z' )
numa[a[i]-'a'] ++;
else num ++ ;
}
for( int i = 0 ; i < lenb ; i++ ){
numb[b[i]-'a'] ++;
}
int l = 0 , r = AX , x = 0;
while( l <= r ){
int mid = (l+r)>>1;
if( check(mid) ) { l = mid + 1 ; x = mid ; }
else r = mid - 1;
}
int id = 0;
for( int i = 0 ; i < lena ; i++ ){
if( a[i] != '?' ) printf("%c",a[i]);
else{
while( id < 26 && numa[id] >= numb[id]*x ) id++;
if( id < 26 ){
printf("%c", id + 'a');
numa[id]++;
}else{
putchar('a');
}
}
}
return 0;
}